Let $\{D_m\}$ be an increasing sequence of simply connected domains, and let $\phi_m$ be the Riemann map of $D_m$ onto the open unit disk $\mathbb{D}$, normalized so that $\phi_m(z_0) = 0$ and $\phi'(z_0) > 0$ for some fixed $z_0 \in D_1$. Let $D$ be the union of the $D_m$'s. Show that if $D$ is the entire complex plane, then the $\phi_m$'s are eventually defined on the disk $\{|z| \leq R\}$ and converge there uniformly to $0$. Otherwise, $D$ is simply connected and the $\phi_m$'s are eventually defined on each compact subset of $D$ and converge there uniformly to the Riemann map $\phi$ of $D$ onto $\mathbb{D}$ satisfying $\phi(z_0) = 0$ and $\phi'(z_0) > 0$.
Obviously I'm missing something here. I feel the first conclusion would always be true. This is map to the unit disk. So wouldn't every point "eventually" (more like right away) be on the disk $\{|z| \leq R\}$ since they are being mapped to the unit disk? What am I missing here and how would I go about this?
Best Answer
The statement you are asking about is
What it means for $\phi'_m$ to be defined on the disc $\{|z| \le R\}$ is that $\{|z| \le R\}$ is a subset of the domain of definition of $\phi'_m$. This can be written more succinctly as $\{|z| \le R\} \subset D_m$, because $D_m$ is the domain of definition of $\phi'_m$ (which is identical to the domain of definition of the Riemann map $\phi_m$).
So, what that statement means is that
Actually this particular statement has nothing to do with $\phi'_m$, and it has nothing to do with simple connectivity: it is true using only the assumption that $D_1 \subset D_2 \subset D_3 \subset \cdots$ is an increasing sequence of open subsets such that $D = \bigcup_{m=1}^\infty D_m = \mathbb C$.
The proof is a simple application of compactness. First, the closed ball $B_R = \{|z| < R\}$ is compact. Next, $\{D_m\}$ is an open cover of $B_R$, because it is an open cover of the larger set $D=\mathbb C$. Applying compactness of $B_R$, there exists a finite subcover $\{D_{m_1},...,D_{m_K}\}$ of $B_R$, meaning such that $B_R \subset D_{m_1} \cup \cdots \cup D_{m_K}$. But we can rearrange the indices to be in increasing order $m_1 < m_2 < ... < m_K$ and it follows that $D_{m_1} \cup \cdots \cup D_{m_K} = D_{m_K}$. So for all $m \ge M = m_K$ it follows that $$D_m \supset D_{m_K} = D_{m_1} \cup \cdots \cup D_{m_K} \supset B_R = \{|z| \le R\} $$